Question

using the quadratic formula to solve x2 = 5 – x, what are the values of x?

Answer 1

The answer will be A. -1+=$\sqrt{21}$/2

Answer 2

Answer:

A quadratic equation $ax^2+bx+c = 0$    ....[1] where, a, b and c are coefficient.

Then the solution is given by:

$x = -\frac{b \pm \sqrt{b^2-4ac}}{2a}$        ....[2]

As per the statement:

Given the equation:

$x^2=5-x$

⇒$x^2+x-5 = 0$

On comparing with [1] we have;

a =1, b = 1 and c = -5

Then substitute these in [2] we have;

$x = \frac{-1 \pm \sqrt{1^2-4(1)(-5)}}{2(1)} =\frac{-1 \pm \sqrt{1+20}}{2} =\frac{-1 \pm \sqrt{21}}{2}$

Therefore, the values of x are:

$x = \frac{-1 + \sqrt{21}}{2}$ , $\frac{-1 - \sqrt{21}}{2}$